Distributional derivative of $\frac{e^{i|x|}}{|x|}$ For $x\in\mathbb{R}^3$ we have $\frac{1}{|x|} \in L^1_{loc}(\mathbb{R}^3)$ and $\frac{1}{|x|} \to 0$ as $|x| \to \infty$. Hence we can conclude that
$$
f(x)=\frac{e^{i|x|}}{|x|}
$$
defines a (tempered) distribution
$$
\phi \mapsto \int_{\mathbb{R}^3}f\phi.
$$
For a distribution $T \in \mathcal{S}'(\mathbb{R}^3)$ its derivative is defined as $(\partial x_i T)\phi = -T(\partial x_i \phi)$.
Now I want to calculate the partial distributional derivative of $f$. Starting with the definition yields
$$
-\int_{\mathbb{R}^3}\frac{e^{i|x|}}{|x|}\partial x_i \phi(x) dx
$$
But how do I proceed? Since $f$ has a singularitie in $0$ there could somewhere appear a delta function.
 A: We have
$$-\int_{\mathbb{R}^3}\frac{e^{i|x|}}{|x|}\partial_j \phi(x) \,dx = -\left ( \lim_{\epsilon \to 0} \int_{\mathbb{R}^3 \setminus B_\epsilon}\frac{e^{i|x|}}{|x|}\partial_j \phi(x) \,dx \right )$$
by dominated convergence. Use the formula for integration by parts in higher dimensions to obtain
$$
\int_{\mathbb{R}^3 \setminus B_\epsilon}\frac{e^{i|x|}}{|x|}\partial_j \phi(x) \,dx = \int_{\partial B_{\epsilon}} \frac{e^{i|x|}}{|x|} \phi(x) \nu_j \,dS(x) - \int_{\Bbb{R}^3\setminus B_{\epsilon}} \phi(x) \partial_j  \frac{e^{i|x|}}{|x|} \,dx.
$$
The integration over the unbounded domain is justified because a surface integral of $f\phi$ over $\partial B_R$ would thend to zero for $\phi \in \mathcal{S}(\mathbb{R}^3)$ and $R \to \infty$.
The first integral on the RHS goes to zero since $|\nu_j| \leq 1$ and the area of the surface of $\partial B_{\epsilon}$ is $4\pi\epsilon^2$.
Further, for $x \neq 0$ we have
$$
\partial_j f(x)=\frac{x_je^{i|x|}(i|x|-1)}{|x|^3},
$$
which has an integrable singularity in $0$. Hence one can conclude that
$$
\partial_j L_f=L_{\partial_jf}.
$$
